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Point Charge on Semicircular Ring

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data

    6ab1703a-9eb8-437e-98ca-897848a9c148.jpg

    A point charge Q is located at point P(0,-4,0) while a 10nc charge is uniformly distributed along a semicircular ring as shown in the figure. find the value of Q such that E(0,0,0)=0


    2. Relevant equations

    Q=ρLdl
    dl = ρd∅ (because ρ and z are constant)
    E=[itex]\frac{Qρ}{4piεr^2}[/itex]

    3. The attempt at a solution
    Okay so first I'd have to find Q using the formula above. I'd set my limits from 0 to pi.

    Now looking at the formula for point charge, the only missing factor is R. I don't know how to express this. I know that the equation of a hemisphere is very similar to that of a circle. I came up with y=[itex]\sqrt{4-x^2}[/itex]. But now that I look at it, if I tried to convert that to polar using the relationships x=rcos[itex]\vartheta[/itex] and y=rsin[itex]\vartheta[/itex], it really seems wrong.

    What's the best procedure in doing this? What should I say to myself every time I see a ring? Because to be honest, I really get confused when I see an odd figure labeled in terms of (x,y,z) even though it's much easier in another system.

    Also, after finding R, I simply just stick to the formula and I will obtain my solution, right?

    Thank you for any help :)
     
  2. jcsd
  3. Oct 18, 2011 #2
    First off you want to treat each case separately. By symmetry of the problem we know that the E-field will only be in the y-direction. Since the charge is distributed uniformly along the arc, you can deal with it as a line charge density which is what you did:
    [itex]dQ_1=\rho_l(\rho d\phi)[/itex] now integrating that out you can get the line charge density in terms of [itex]Q_1[/itex] (which you'll want in the end).

    So setting up the integral for the E-field:
    Since your r is constant (r=2) you'll only be integrating over the angular change.
    [itex]dE_y=\frac{\rho_l}{2\pi\epsilon_or}sin\phi d\phi[/itex]

    Integrating that out should give you the E-field in the y-direction due to the hemisphere.

    The point charge is just the E-field due to a point charge which is:
    [itex]E = \frac{Q}{4\pi\epsilon_oR^2}[/itex] where Q is the charge you're solving for and R is the distance to the charge.

    The total E-field:
    [itex]E_t = E_{ring}-E_{charge}[/itex] and just solve for Q.

    Hope this helps.
     
  4. Oct 18, 2011 #3
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